Answer:
[tex]\large \boxed{1 \,^{\circ}\text{C}}[/tex]
Explanation:
There are two heat transfers involved: the heat gained by the spoon and the heat lost by the coffee.
According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.
Let the spoon be Component 1 and the coffee be Component 2.
Data:
For the spoon:
[tex]m_{1} =\text{45 g; }T_{i} = 25 ^{\circ}\text{C; }\\C_{1} = 0.24 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}[/tex]
For the coffee:
[tex]m_{2} =\text{180 g; }T_{i} = 95 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}[/tex]
Calculations
(a) The relative temperature change
[tex]\begin{array}{rcl}\text{Heat gained by spoon + heat lost by coffee} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{45 g}\times 0.24 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{180 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\10.8\Delta T_{1} + 753.1\Delta T_{2} & = & 0\\753.1\Delta T_{2} & = & -10.8\Delta T_{1}\\\Delta T_{2} & = & -0.01434\Delta T_{1}\\\end{array}[/tex]
The ΔT for the coffee is about 1/70 that of the spoon.
(b) Final temperature of coffee
[tex]\Delta T_{1} = T_{\text{f}} - 25 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 95 ^{\circ}\text{C}[/tex]
[tex]\begin{array}{rcl}\Delta T_{2} & = & -0.01434\Delta T_{1}\\T_{\text{f}} - 95 ^{\circ}\text{C} & = & -0.01434 (T_{\text{f}} - 25 ^{\circ}\text{C})\\& = & -0.01434T_{\text{f}} + 0.3585 ^{\circ}\text{C}\\T_{\text{f}} & = & -0.01434T_{\text{f}}+ 95.36^{\circ}\text{C}\\1.01434T_{\text{f}}& = & + 95.36 ^{\circ}\text{C}\\T_{\text{f}} & = & \dfrac{ 95.36 ^{\circ}\text{C}}{1.01434}\\\\ & = & \mathbf{94.0 ^{\circ}}\textbf{C}\\\end{array}\\[/tex]
(c) Decrease in coffee temperature
[tex]\Delta T_{2} = T_{\text{f}} - 95 ^{\circ}\text{C} =94.0 ^{\circ}\text{C} -95^{\circ}\text{C}=-1 \,^{\circ}\text{C}\\\text{The temperature of the coffee will be reduced by $\large \boxed{\mathbf{ 1 \,^{\circ}\textbf{C}}}$}[/tex]
